Number of divisors of a number $30^{30}$ which have 30 divisors
The task is to calculate the number of divisors of a number $30^{30}$, but only of those which have 30 divisors themselves.
Basically, I can count how many divisors given number has, since they are all in a form $2^\alpha3^\beta5^\gamma$, such that $\alpha,\beta,\gamma\leqslant30$.
What bothers me is how to approach to the other part of the problem.
Any hints would be appreciated.
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$\begingroup$$\mathbf{Hint:}$ We see that there are $(31)^3$ divisors of this number. Now we want the numbers which themselves have 30 divisors. The prime factors available to us are 2,3 and 5. So we find the number of ways in which 30 can be broken into product of three factors and then after subtracting 1 from each factor arrange them appropriately as the powers of 2, 3 and 5 respectively. e. g. 30 can be broken as $1.1.30$ so we can write the original divisor in 3 ways as $2^0.3^0.5^{29}$, $2^0.3^{29}.5^0$ and $2^{29}.3^0.5^0$. Some other ways to break this 30 are $(2.3.5 ; 1.2.15 ; 1.3.10 ; 1.5.6)$. Thus the answer may be $3+3!+3!+3!+3!$ i.e. $\mathbf{27}$
$\endgroup$ $\begingroup$$30^{30} = 2^{30}*3^{30}*5^{30}$.
Every divisor will be of the form $2^a3^b4^c$ where each $a,b$ or $c$ may be any integer between $0$ and thirty. So $30^{30}$ has $31^3$ divisors.
Now you want only the divisors with $30$ divisors. $2^a3^b5^c$ will have divisors of the form $2^{\alpha}3^{\beta}5^{\gamma}$ where $0 \le \alpha \le a;0 \le \beta \le b; 0 \le \gamma \le c$. There are by $(a+1)(b+1)(c+1)$ divisors. If $2^a3^b5^c$ has $30$ divisors $(a+1)(b+1)(c+1)=30$.
So .... how many ways are there to choose $a,b,c$ so that $0 \le a,b,c \le 30$ and $(a+1)(b+1)(c+1) = 30$?
If $c=a+1; d=b+1;e=c+1$ we can have $(c,d,e) = (30,1,1), (15,2,1), (15,2,1) ..etc. $ So we can have the divisors $2^{29}, 2^{14}*3, 2^{14}*5$... etc.
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